**Write the first terms of the sequences whose nth terms is a**_{n}= n(n + 2)

Solution:

a_{n} = n(n + 2)

Substituting n = 1, 2, 3, 4 and 5 we obtain

a_{1} = 1(1+2) = 3

a_{2} = 2(2+2) = 8

a_{3} = 3(3+2) = 15

a_{4} = 4(4+2) = 24

a_{5} = 5(5+2) = 35

Therefore, the required terms are 3, 8, 15, 24 and 35.

**Write the first five terms of the sequences whose n**^{th}terms is a_{n}=^{n}/_{(n+1)}

Solution:

a_{n} = ^{n}/_{(n+1)}

Substituting n = 1, 2, 3, 4, 5 obtain

a_{1} = ^{1}/_{1+1} = ^{1}/_{2} ,

a_{2} = ^{2}/_{2+1} = ^{2}/_{3} ,

a_{3} = ^{3}/_{3+1} = ^{3}/_{4} ,

a_{4} = ^{4}/_{4+1} = ^{4}/_{5} ,

a_{5} = ^{5}/_{5+1} = ^{5}/_{6} ,

Therefore, the required terms are ^{1}/_{2}, ^{2}/_{3} ,^{ 3}/_{4} ,^{ 4}/_{5} and ^{5}/_{6}

**Write the first five terms of the sequences whose n**^{th}term is a_{n}= 2^{n}

Solution:

a_{n} = 2^{n}

Substituting n = 1, 2, 3, 4, 5 we obtain

a_{1} = 2^{1} = 2

a_{2} = 2^{2} = 4

a_{3} = 2^{3} = 8

a_{4} = 2^{4} = 16

a_{5} = 2^{5} = 32

Therefore the required terms are 2, 4, 8, 16 and 32

**Write the first five terms of the sequences whose n**^{th }terms is a_{n}=^{(2n-3)}/_{6}

Solution:

Substituting n = 1, 2, 3 , 4, 5 we obtain

a_{1} = ^{(2×1-3)}/_{6} = ^{-1}/_{6 },

a_{2} = ^{(2×2-3)}/_{6} = ^{1}/_{6 },

a_{3} = ^{(2×3-3)}/_{6} = ^{3}/_{6} = ^{1}/_{2 },

a_{4} = ^{(2×4-3)}/_{6} = ^{5}/_{6 },

a_{5} = ^{(2×5-3)}/_{6} = ^{7}/_{6 },

Therefore, the required terms are ^{-1}/_{6 }, ^{1}/_{6 },^{ 1}/_{2 },^{ 5}/_{6 } and ^{7}/_{6 }.

**Write the first five terms of the sequences whose n**^{th}terms is a_{n}= (-1)^{n-1}5^{n+1}

Solution:

Substituting n = 1, 2, 3, 4, 5 we obtain

a_{1} = (-1)^{1-1} 5^{1+1} = 5^{2} = 25

a_{2} = (-1)^{2-1} 5^{2+1} = -5^{3} = -125

a_{3} = (-1)^{3-1} 5^{3+1} = 5^{4} = 625

a_{4} = (-1)^{4-1} 5^{4+1} = -5^{5} = -3125

a_{5} = (-1)^{5-1} 5^{5+1} = 5^{6} = 15625

Therefore, the required terms are 25, -125, 625, -3125 and 15625.

**Write the first five terms of the sequences whose n**^{th }term is a_{n}= n[(n^{2}+5)/_{4}]

Solution:

Substituting n = 1, 2, 3, 4, 5 we obtain

a_{1} = 1. [(1^{2}+5)/_{4}] = ^{6}/_{4} = ^{3}/_{2},

a_{2} = 2. [(2^{2}+5)/_{4}] = 2. ^{9}/_{4} = ^{9}/_{2},

a_{3} = 3. [(3^{2}+5)/_{4}] = 3. ^{14}/_{4} = ^{21}/_{2},

a_{4} = 4. [(4^{2}+5)/_{4}] = 21,

a_{5} = 5. [(5^{2}+5)/_{4}] = 5. ^{30}/_{4} = ^{75}/_{2},

Therefore, the required terms are ^{3}/_{2},^{ 9}/_{2}, ^{21}/_{2}, 21 and^{ 75}/_{2}.

**Find the 17**^{th}in the following sequences whose n^{th}term is a_{n}= 4n – 3 , a_{17}, a_{24}

Solution:

Substituting n = 17 we obtain

a_{17} = 4(17) – 3 = 68 – 3 = 65

Substituting n = 24 we obtain

a_{24} = 4(24) – 3 = 96 – 3 = 93

**Find the 7**^{th}term in the following sequences whose nth terms is a_{n}= [(n^{2})/_{2n}]; a_{7}

Solution:

Substituting n = 7 we obtain

a_{7} = (7^{2})/(2^{7}) = ^{49}/_{128}

**Find the 9**^{th}term in the following sequences whose nth term is a_{n}= (-1)^{n-1}n^{3}; a_{9}

Solution:

Substituting n = 9 we obtain

a_{9} = (-1)^{9-1}(9)^{3} = (9)^{3} = 729

**Find the 20**^{th}term in the following sequences whose nth terms is a_{n}= [{n(n+2)}/_{(n+3)}]; a_{20}

Solution:

Substituting n = 20, we obtain

a_{20} = [{20(20+2)}/_{(20+3)}] = ^{20(18)}/_{23} = ^{360}/_{23}

**Write the first five terms of the following sequences and obtain the corresponding series:**

**a _{1} = 3, a_{n} = 3a_{n-1} + 2, for all n > 1**

Solution:

a_{1} = 3, a_{n} = 3a_{n-1} + 2 for all n>1

a_{2} = 3a_{1} + 2 = 3(3) + 2 = 11

a_{3} = 3a_{2} + 2 = 3(11) + 2 = 35

a_{4} = 3a_{3} + 2 = 3(35) + 2 = 107

a_{5} = 3a_{4} + 2 = 3(107) + 2 = 323

Hence the first five terms of the sequences are 3, 11, 35, 107 and 323.

The corresponding series 3 + 11+ 35 + 107 + 323 + …

**Write the first five terms of the following sequence and obtain the corresponding series:**

**a _{1} = -1, a_{n} = [(a_{n-1})/_{n}], n ≥ 2**

Solution:

a_{1} = -1, a_{n} = [(a_{n-1})/_{n}], n ≥ 2

a_{2} = (a_{1})/_{2} = ^{-1}/_{2},

a_{3} = (a_{2})/_{3} = ^{-1}/_{6},

a_{4} = (a_{3})/_{4} = ^{-1}/_{24},

a_{5} = (a_{4})/_{4} = ^{-1}/_{120},

Hence the first five terms of the sequences are -1, ^{-1}/_{2}, ^{-1}/_{6},^{ -1}/_{24},^{ -1}/_{120},

The corresponding series is (-1)+(^{ -1}/_{2})+(^{ -1}/_{6})+(^{ -1}/_{24})+(^{ -1}/_{120})+…

**Write the first five terms of the following sequence and obtain the corresponding series:**

**a _{1} = a_{2} = 2, a_{n} = a_{n-1} -1, n > 2**

Solution:

a_{1} = a_{2} = 2, a_{n} = a_{n-1} -1, n > 2

a_{3} = a_{2} – 1 = 2 – 1 = 1

a_{4} = a_{3} – 1 = 1 – 1 = 0

a_{5} = a_{4} – 1 = 0 – 1 = -1

Hence, the first five terms of the sequence are 2, 2, 1, 0 and -1 and the corresponding series is 2 + 2 + 1 + 0 + (-1) + …

**The Fibonacci sequence is defined by 1 = a**_{1}= a_{2}and a_{n}= a_{n-1}+ a_{n-2}, n>2

**Find [(a _{n+1})/(a_{n})], for n = 1, 2, 3, 4, 5**

Solution:

1 = a_{1}= a_{2}

a_{n} = a_{n-1} + a_{n-2}, n>2

a_{3} = a_{2} + a_{1} = 1 + 1 = 2

a_{4} = a_{3} + a_{2} = 2 + 1 = 3

a_{5} = a_{4} + a_{3} = 3 + 2 = 5

a_{6} = a_{5} + a_{4} = 5 + 3 = 8

For n = 1, [(a_{n+1})/(a_{n})] = [(a_{2})/(a_{1})] = ^{1}/_{1} = 1

For n = 2, [(a_{n+1})/(a_{n})] = [(a_{3})/(a_{2})] = ^{2}/_{1} = 2

For n = 3, [(a_{n+1})/(a_{n})] = [(a_{4})/(a_{3})] = ^{3}/_{2}

For n = 4, [(a_{n+1})/(a_{n})] = [(a_{5})/(a_{4})] = ^{5}/_{3}

For n = 5, [(a_{n+1})/(a_{n})] = [(a_{6})/(a_{5})] = ^{8}/_{5}